Research on fault detection for Markovian jump systems with time-varying delays and randomly occurring nonlinearities

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1 DOI: /CTA Control Theory & Applcatons Vol. 33 No. 9 Sep ( ; ) (RONs)... H. Lyapunov H.. ; ; ; TP273 A Research on fault detecton for Markovan jump systems wth tme-varyng delays and randomly occurrng nonlneartes REN We-jan 1 WANG Chun-le 1 LU Yang 2 (1. College of Electrc and Informaton Engneerng Northeast Petroleum Unversty Daqng Helongjang Chna; 2. College of Informaton Technology Helongjang Bay Agrcultural Unversty Daqng Helongjang Chna) Abstract: Ths paper s concerned wth the fault detecton problem for a class of dscrete-tme Markovan jump systems wth multple tme-varyng stochastc delays and randomly occurrng nonlneartes. Accordng to the stochastc characterstc of multple tme-varyng delays an ndvdual Bernoull dstrbuton s used to descrbe ths phenomenon. Furthermore a knd of stochastc nonlnear dsturbance s also consdered n the fault detecton research by means of a Bernoull dstrbuted whte sequence. The man purpose of ths paper s to desgn a fault detecton flter such that the resdual system s mean square exponental stable and the relevant H performance s satsfed. Through Lyapunov stablty theory and stochastc analyss technology suffcent condtons are set up for the desred exponental stablty and H dsturbance attenuaton a convex optmzaton problem s solved by the semdefnte programmng technque the gan characterstc of the fault detecton flter s obtaned. Fnally a smulaton example s provded to llustrate the usefulness and effectveness of the proposed desgn approach. Key words: fault detecton; Markovan jump system; stochastc tme-varyng delays; randomly occurrng nonlneartes 1 (Introducton)... [1. (Markovan jump systems MJSs) [2. MJSs. MJSs [3 4 [5 11 [12. MJSs ; E-mal: wangcl001@163.com; Tel.: ( ) (LBH Q12143) (QC2013C066). Supported by Natonal Natural Scence Foundaton of Chna ( ) Helongjang Postdoctoral Research Foundaton (LBH Q12143) Helongjang Youth Foundaton (QC2013C066).

2 [13 [14 ; [15. [ [ [ [24.. 1). 2). 3). 4) Lyapunov H. 2 ; 3 Lyapunov LMI ; 4 ; M T M ; R n n R n m n m ; Z ; I 0 ; l 2 ([0 ); R n ) n [0 ) ; P>0 P ; E{x} x ; x x ; dag{a 1 A 2 A n } A 1 A 2 A n *. 2 (Problem formulaton) x(k +1)=A(θ k )x(k)+γ(k)g(θ k x(k))+ A d1 (θ k ) q α r (k)x (k τ r (k)) + D 1 (θ k )ω(k)+g(θ k )f(k) y(k) =C(θ k )x(k)+d 2 (θ k )ω(k)+ E(θ k )f(k) x(k) =ψ(k) k Z (1) x(k) R n ; y(k) R m ; ω(k) R p l 2 [0 ) ; ( f(k) l ) 2 [0 );R l. A(θ k ) A d1 (θ k ) G(θ k ) C(θ k ) D 1 (θ k ) D 2 (θ k ) E(θ k ). τ r (k)(r =1 2 q) ; ψ(k) ; θ k S = {1 2 N} p j =P{θ k+1 = j θ k = } N θ j =1. j=1 g(θ k x(k)) g(θ k 0) = 0 g(θ k x(k)+δ(k)) g(θ k x(k)) B 1 (θ k )δ(k) (2) B 1 (θ k ) δ(k). α r (k)(r =1 2 q) γ(k) 0 1 (RONs) P{α r (k) =1} =ᾱ r P{α r (k) =0} =1 ᾱ r (3) P{γ(k) =1} = γ P{γ(k) =0} =1 γ ᾱ r [0 1 γ [ g(θ k x(k)). γ(k) g(θ k x(k)) (RONs) RONs. τ r (k)(r =1 2 q)

3 d m τ r (k) d M d m d M. 3 1) 3 ; 2).. A(θ k )=A A d1 (θ k )=A d1 g(θ k x(k)) = g (x(k)) G(θ k )=G C(θ k )=C D 1 (θ k ) = D 1 D 2 (θ k )=D 2 E(θ k )=E B 1 (θ k )=B 1. ˆx(k +1)=A F (θ k )ˆx(k)+B F (θ k )y(k) (4) r(k) =C F (θ k )ˆx(k)+D F (θ k )y(k) ˆx k R n r(k) R l A F (θ k ) B F (θ k ) C F (θ k ) D F (θ k ). A F (θ k )= A F B F (θ k )=B F C F (θ k )=C F D F (θ k )=D F. (1) (4) x(k +1)=Ā(θ k) x(k)+ q (Ādr(θ k )+ Ã dr (θ k )) x(k τ r (k))+ ( γ + γ(k))zg(θ k x(k))+ (5) D(θ k )v(k) r(k) = C(θ k ) x(k)+ D F (θ k )v(k) x(k) = [ x T (k) ˆx T (k) T r(k) =r(k) f(k) v(k)= [ ω T (k) f T (k) [ T A 0 Ā(θk )= B F C A F [ᾱr A d1 0 [ αr (k)a d1 0 Ā dr (θ k )= Ã dr (θ k )= D1 G D(θ k )=[ Z=[I 0 T B F D 2 B F E C(θ k )=[D F C C F D F (θ k )=[D F D 2 D F E I α r (k) =α r (k) ᾱ r γ(k)=γ(k) γ. E{ α r (k)} =0 E{ α r(k)} 2 =ᾱ r (1 ᾱ r ) E{ γ(k)} =0 E{ γ 2 (k)} = γ(1 γ). Ā(θ k)=ā Ā dr (θ k )=Ādr Ã dr (θ k ) = Ãdr D(θk )= D C(θ k )= C D F (θ k )= D F. 1 v(k) =0 δ >0 0 <κ<1 E{ x(k) 2 } δ k κ sup Z E{ ψ() 2 } k 0 (5).. (4) A F B F C F D F a) (5). b) γ > 0 v(k) r(k) E{ r(k) 2 } γ 2 E{ v(k) 2 } (6) γ >0 (6). J(k) J th : J(k) ={ L r T (h)r(h)} 1 2 h=0 (7) J th = sup E{J(k)} ω k l 2f k =0 L. (7). J(k) >J th J(k) J th. 3 (Man results) Lyapunov (1).. 1 ( Schur [25.) S 1 S 2 S 3 S 1 = S1 T 0 <S 2 = S2 T S 1 + S3 T S 1 2 S 3 < 0 [ [ S1 S3 T S2 S 3 < 0 < 0. (8) S 3 S 2 S3 T S 1 2 x R n y R n Q >0 x T Qy + y T Qx x T Qx + y T Qy. (6). 1 (5) H γ >0 P > 0( N) Q r > 0(r =1 2 3 q) ρ >0

4 Φ 1 + C T C 2 γ x T Φ = Ẑ T P Zg (x(k))+ P Ā Φ 2 < 0 q D T P Ā + D F T C x DT P Ẑ Φ (k τ r (k))ãt P dr Ã dr x(k τ r (k))+ 3 (9) 2( q Ā dr x(k τ r (k))) T P D v(k)+ Z T P Z ρi (10) v T (k) P D v(k)+ (6) 2 γ( q Ā dr x T (k τ r (k))) T P Zg (x(k))+ P = N p j P j γg T (x(k))z T P Zg (x(k)) j=1 (12) Φ 1 =2ĀT P Ā + q (d M d m +1)Q r P + ρ B dr Ã dr } = Φ 2 =dag{ Q 1 + Ã1 Q 2 + Ã2 Q q + [ T [ αr (k)a d1 0 αr (k)a d1 0 Ã q } +2ẐT P Ẑ E{ P } = Φ 3 =2 D T P D + D T D F F γ 2 I [ T [ Ã r =ᾱ r (1 ᾱ r )ÂT P Ad1 0 Ad1 0 d1 Â d1 2 q ᾱ r (1 ᾱ r ) P = Ẑ =[Ād1 Ā d2 Ā dq ᾱ B =dag{(3 γ 2 + γ)b T r (1 ᾱ r B 0} d1 Â d1. (13) 2 Â d1 =dag{a d1 0}. 2 γ x T P Zg (x(k)) η(k) := [ x T (k) x T (k τ 1 (k)) x T (k τ q (k)) T x T. P Ā x(k)+ γ 2 g T (x(k))z T P Zg (x(k)) (14) θ k = N 2 γg T (x(k))z T P D v(k) Lyapunov-Krasovsk γ 2 g T (x(k))z T P Zg (x(k)) + v T (k) D T P D v(k) V ( x(k)θ k )= (15) V 1 ( x(k)θ k )+V 2 ( x(k)θ k )+V 3 ( x(k)θ k ) (11) 2 γ( q Ā dr x(k τ r (k))) T P Zg (x(k)) ( q Ā dr x(k τ r (k))) T P ( q Ā dr x(k τ r (k)))+ V 1 ( x(k)θ k )= x T (k)p (θ k ) x(k) V 2 ( x(k)θ k )= q k 1 x T (l)q r x(l) V 3 ( x(k)θ k )= q l=k τ r (k) d m k 1 m= d M+1 l=k+m x T (l)q r x(l). ΔV 1 ( x(k)θ k )= E {V 1 ( x(k +1)θ k+1 ) x(k)θ k = } V 1 ( x(k)θ k )= N p j x T (k +1)P j x(k +1) x T (k)p x(k) = j=1 x T (k)(āt P Ā P ) x(k)+ 2 x T (k)āt P ( q Ā dr x(k τ r (k)))+ 2 x T (k)āt P D v(k)+2 γg T (x(k))z T P D v(k)+ ( q Ā dr x(k τ r (k))) T P ( q Ā dr x(k τ r (k)))+ γ 2 g T (x(k))z T P Zg (x(k)). (16) (2) g T (x(k))(3 γ 2 + γ)z T P Zg (x(k)) x T (k)(3 γ 2 + γ)ρb T B x(k) = x T (k)ρ B x(k) (12) (17) ΔV 1 ( x(k)θ k )= E {V 1 ( x(k +1)θ k+1 ) x(k)θ k = } V 1 ( x(k)θ k ) x T (k)(2āt P Ā P + ρ B ) x(k)+ 2 x T (k)āt P ( q Ā dr x(k τ r (k)))+ 2 x T (k)āt (17) P D v(k)+2( q Ā dr x(k τ r (k))) T

5 P ( q Ā dr x(k τ r (k)))+ k d m l=k d M+1 x T (l)q r x(l)) (21) q x T (k τ r (k))ãt P dr Ã dr x(k τ r (k))+ v(k) =0 2( q E{ΔV ( x(k)θ k )} η T (k)φ η(k) (22) Ā dr x(k τ r (k))) T P D v(k)+ 2v T (k) D [ T P D v(k) (18) Φ1 Φ =. ΔV 2 ( x(k)θ k )= Ẑ T P Ā Φ 2 E {V 2 ( x(k +1)θ k+1 ) x(k)θ k = } V 2 ( x(k)θ k )= 1 Φ < 0. (5). N p j ( q k x T (5) H (l)q r x(l) j=1 l=k τ r (k)+1 q k 1 x T (l)q r x(l)) J N =E [ r T (k) r(k) β 2 v T (k)v(k) = l=k τ r(k) q ( x T E [ r (k)q r x(k) (k) r(k)+δv( x(k)θ k ) x T β 2 v T (k)v(k) E {V (k τ r (k))q r x(k τ r (k))+ k+1 ( x(k+1)θ k+1 )} k d m E [ r T (k) r(k) β 2 v T (k)v(k)+ x T (l)q r x(l)) (19) l=k d M +1 ΔV ( x(k)θ k ) ΔV 3 ( x(k)θ k )= E {V 3 ( x(k +1)θ k+1 ) x(k)θ k = } η T (k)φη(k) < 0. V 3 ( x(k)θ k )= q ((d M d m ) x T (k)q r x(k) E{ r(k) 2 } γ 2 E{ v(k) 2 }. k d m x T (l)q r x(l)). (20). l=k d M +1. (18) (20) 2 ΔV ( x(k)θ k ) x T (k)(2āt P Ā P + ρ B ) x(k)+. H γ >0. P > 0( N) Q r > 0 2 x T (k)āt P ( q (r =1 2 3 q) ρ>0 X K Ā dr x(k τ r (k)))+ 2 x T (k)āt P D v(k)+2( q Ā dr x(k τ r (k))) T Y = ˆΓ P ( q 11 Ā dr x(k τ r (k)))+ 0 γ 2 I q x T (k τ r (k))ãt P ˆΓ dr Ã dr x(k τ r (k))+ 31 P ˆD0 + X ˆR2 P < 0 ˆΓ 41 P 2( q ˆΓ Ā dr x(k τ r (k))) T 51 ˆΓ52 ˆP P D v(k)+ (23) 2v T (k) D T P D v(k)+ q ( x T (k)q r x(k) Z T P Z ρi (24) x T (4) (k τ r (k))q r x(k τ r (k))+. k d m l=k d M +1 x T (l)q r x(l))+ q ((d M d m ) x T (k)q r x(k) ˆΓ 11 =dag{γ 11 Γ 22 } ˆΓ 31 =[ P Â 0 + X ˆR1 P Ẑ ˆΓ 41 =dag{ P Â 0 + X ˆR1 P Ẑ }

6 Γ 11 = ρ B + q (d M d m +1)Q r P. Γ 22 =dag{ Q 1 + Ã1 Q 2 + Ã2 [ Q q + [ Ãq} ˆΨ = ˆΓ 51 = K ˆR1 0 [ [ P ˆD0 + X ˆR2 0n n ˆΓ 52 = Ê = [ [ K ˆR2 ÊT 1 I n n P =dag { P P } ˆP =dag { P I } A 1 = A 2 =.7.5 [ Ê 1 =[0 l p I l p T Ê 2 =[0 m n I m n T A [ [ d11 = A 0 D1 G [ Â 0 = ˆD0 = A [ [ d12 = I [ [ ˆR 1 = ˆR2 =. C 0 D 2 E D 11 = D 12 =.3.4 (P Q r X K ρ) (23) (24) (4) C 1 = C 2 =[ H 1 = H 2 =0.9 D 21 =[ D 22 =[ [A F B F =[ÊT P Ê 1 Ê T X G (25) 1 =[ T G 2 =[ T. [C F D F =K. 2 τ r (k) 6(r =1 2) P Q r 1. ᾱ 1 =E{α 1 (k)} =0.9 Ā = Â0 + ÊK ˆR 1 1 ᾱ 2 =E{α 2 (k)} =0.7 D = ˆD 0 + ÊK ˆR 1 2 (26) γ =E{γ(k)} =0.8. C = K ˆR1 D F = K ˆR2 ÊT 1 K 1 =[A F B F. Shur (9) ˆΓ 11 0 γ 2 I ˆΓ 31 ˆD0 +ÊK ˆR 1 2 P 1 1 ˆΓ 41 P ˆΓ 51 ˆΓ52 ˆP 1 ˆΓ 31 =[Â0 + ÊK ˆR 1 1 Ẑ ˆΓ 41 =dag{â0 + ÊK ˆR 1 1 [ Ẑ} ˆD0 + ˆΓ 52 = ÊK ˆR 1 2. K ˆR2 ÊT 1 <0 (27) 1 X = P ÊK 1 (23). 4 (Numercal example) g 1 (x(k)) = g 2 (x(k)) = 0.5x 1 (k)sn(x 2 (k)) ε(k) =1 B(k) =dag{ } (2). 2 (25) [ A F1 = B F1 = [ C F1 = D F1 = [ A F2 = B F2 = [ [ C F2 =[ D F2 = k = f(k) { 1 40 k 80 f(k) = 0. ˆΨ θ(0) = 1 θ(k) 1.

7 { [rand[ rand[0 1 T 0 k 50 ω(k) = 0 1 Fg. 1 Modes evoluton ω(k) =0 r(k) J(k) 2 3. rand [0 1. r(k) J(k) 4 5. J th =sup f=0 r T (s)r(s)} 1/2 E{ 150 s=0 200 J th =J(40)<J th <J(41) = Fg. 4 Resdual sgnal wth dsturbance 2 Fg. 2 Resdual sgnal wthout dsturbance 5 Fg. 5 Resdual evaluaton functon wth dsturbance 3 Fg. 3 Resdual evaluaton functon wthout dsturbance 5 (Conclusons)..

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